82 TECHNICAL SURVEY 
effective and the electric intensity is reduced to 
one-half, considering the unobstructed wave as unity. 
Outside the edge, Figure 18, at a distance ab the 
electric intensity is that due to the half of the 
wave, plus such portions of the zones between a and 
b that are uncovered. If an even number of zones is 
uncovered there is approximately a minimum of 
radiation received at the line a, that is, the half 
wave plus the effect of the two zones, 4% + m — mz, 
for the case shown. If a were moved to the right so 
that slightly less than one zone were uncovered there 
would be a maximum, 14 + mi, in which case m, is 
greater than one-half owing to the partial screening 
of the other zones, which, if allowed to operate, 
would reduce the effect due to the-right-hand haif 
of the central zone. For this reason the fringes 
formed outside the shadow may exceed the electric 
intensity of the unobstructed wave. As a is moved 
to the left, more zones are uncovered, and the 
maxima and minima are spaced approximately 
according to the radii of the zones; that is, the 
distances are proportional to the square roots of 
1, 2, 3, ete. : 
Fresnel Integrals 
The preceding discussion is approximate and 
provides a qualitative picture of diffraction phenom- 
ena. The problem will now be formulated quanti- 
tatively by the method of Fresnel. Since the applica- 
tions in view all have to do with diffraction by 
straight edges, slits, etc., the theoretical approach 
will be limited to diffraction of cylindrical waves by 
long edges parallel to the axis of the cylinder. The 
diffraction images of the source will then be bright 
bands also parallel to this axis, and the whole prob- 
lem may be reduced to the consideration of rays in 
a plane perpendicular to the axis. The fact that in 
the applications to be discussed later the illumina- 
tion is due to a point source rather than a line source 
is probably of little importance provided the distance 
from the source to the diffracting edge is sufficiently 
large. 
In Figure 19 is shown a cylindrical wavefront AB 
with its axis at the line source S’ (say an illuminated 
narrow slit). The secondary wavelets from the 
various line elements ds of the wavefront arrive at 
P with different phases, having traveled different 
distances MP. It is desired to find the resultant 
field strength MP due to wavelets from any given 
finite part of the front. 
Let the electric field strength at a point in the 
wavefront be given by the expression 
E = Eysin 2zft , (17) 
where ¢ is the time, f the frequency, and Ep the' 
amplitude of HZ. The phase has been adjusted so as 
to make H = 0 whent = 0. 
Ficure 19. Effect at point P of wavefront AB. 
Consider next the secondary wavelets spreading 
from the front in the direction of P. The field 
intensity at P due to the secondary wavelet emanat- 
ing from the line element ds at the point M (see 
Figure 19) is proportional to ds#» and is inversely 
proportional to the square root of the distance 
MP = d (since this is a cylindrical wave). Further, 
the field intensity must show a phase retardation 
corresponding to the distance d, that is 2rd/y. Hence 
the field strength of the wavelet at P is given by 
an expression of the form 
dE = kEods sin (2aft — = ), (18) 
where k is a factor of proportionality which depends 
to some degree on the angle IPM, and the distance 
d, but which will be considered constant here, as the 
dependence of the phase on d is of much greater 
importance. To obtain the intensity due to wavelets 
emanating from a finite part of the front, equation 
(18) must be integrated over the corresponding 
region of s. For this purpose we need a relation 
between d and s. This is obtained by applying the 
cosine law to the triangle MSP, which gives at once 
d? = (a + b)? + a? — 2a (a + b) cos =, (19) 
or after a simple reduction, using the identity 
Spoon & 
cos (s/2) = il 2 sin Da 
then 5 
d? = b? + 4a (a + b) sin? a: (20) 
For the present purpose it is sufficient to consider 
the case when angle s/a is so small that powers of 
s/a above the square may be neglected in comparison 
with unity. This means that 
ERS 
= Py 
d= ye tot nome b 
(@aeO) oan 8 
A 20a sin” 55 b+ ——— 
or again, on writing 
